Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $q = \dfrac{a^2 + 12a + 27}{10a^2 + 30a} \div \dfrac{2a + 18}{-9a + 54} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{a^2 + 12a + 27}{10a^2 + 30a} \times \dfrac{-9a + 54}{2a + 18} $ First factor the quadratic. $q = \dfrac{(a + 9)(a + 3)}{10a^2 + 30a} \times \dfrac{-9a + 54}{2a + 18} $ Then factor out any other terms. $q = \dfrac{(a + 9)(a + 3)}{10a(a + 3)} \times \dfrac{-9(a - 6)}{2(a + 9)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (a + 9)(a + 3) \times -9(a - 6) } { 10a(a + 3) \times 2(a + 9) } $ $q = \dfrac{ -9(a + 9)(a + 3)(a - 6)}{ 20a(a + 3)(a + 9)} $ Notice that $(a + 3)$ and $(a + 9)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ -9\cancel{(a + 9)}(a + 3)(a - 6)}{ 20a(a + 3)\cancel{(a + 9)}} $ We are dividing by $a + 9$ , so $a + 9 \neq 0$ Therefore, $a \neq -9$ $q = \dfrac{ -9\cancel{(a + 9)}\cancel{(a + 3)}(a - 6)}{ 20a\cancel{(a + 3)}\cancel{(a + 9)}} $ We are dividing by $a + 3$ , so $a + 3 \neq 0$ Therefore, $a \neq -3$ $q = \dfrac{-9(a - 6)}{20a} ; \space a \neq -9 ; \space a \neq -3 $